Zero Exponent Rule: Using the laws of exponents

We use the formula:

$a^0=1$

$\frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3$

We know that:

$9=3^2$

Therefore, we obtain:

$3^2\times3=3^2\times3^1$

We use the formula:

$a^m\times a^n=a^{m+n}$

$3^2\times3^1=3^{2+1}=3^3$

When raising any number to the power of 0 it results in the value 1, mathematically:

$X^0=1$

Apply this to both the numerator and denominator of the fraction in the problem:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1\cdot6^7}{36^4\cdot1}=\frac{6^7}{36^4}$

Note that -36 is a power of the number 6:

$36=6^2$

Apply this to the denominator to obtain expressions with identical bases in both the numerator and denominator:

$\frac{6^7}{36^4}=\frac{6^7}{(6^2)^4}$

Recall the power rule for power of a power in order to simplify the expression in the denominator:

$(a^m)^n=a^{m\cdot n}$

Recall the power rule for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

Apply these two rules to the expression that we obtained above:

$\frac{6^7}{(6^2)^4}=\frac{6^7}{6^{2\cdot4}}=\frac{6^7}{6^8}=6^{7-8}=6^{-1}$

In the first stage we applied the power of a power rule and proceeded to simplify the expression in the exponent of the denominator term. In the next stage we applied the second power rule - The division rule for terms with identical bases, and again simplified the expression in the resulting exponent.

Finally we'll use the power rule for negative exponents:

$a^{-n}=\frac{1}{a^n}$

We'll apply it to the expression that we obtained:

$6^{-1}=\frac{1}{6}$

Let's summarize the various steps of our solution:

$\frac{4^0\cdot6^7}{36^4\cdot9^0}=\frac{1}{6}$

Therefore the correct answer is A.

We use the formula:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

We decompose the fraction inside of the parentheses:

$(\frac{1}{7})^4=\frac{1^4}{7^4}$

We obtain:

$7^4\times8^3\times\frac{1^4}{7^4}$

We simplify the powers: $7^4$

We obtain:

$8^3\times1^4$

Remember that the number 1 in any power is equal to 1, thus we obtain:

$8^3\times1=8^3$

We'll use the law of exponents for negative exponents, but in the opposite direction:

$\frac{1}{a^n} =a^{-n}$Let's apply this law to the problem:

$4^5-4^6\cdot\frac{1}{4}= 4^5-4^6\cdot4^{-1}$When we apply the above law to the second term from the left in the sum, and convert the fraction to a term with a negative exponent,

Next, we'll use the law of exponents for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Let's apply this law to the expression we got in the last step:

$4^5-4^6\cdot4^{-1} =4^5-4^{6+(-1)}=4^5-4^{6-1}=4^5-4^{5}=0$When we apply the above law of exponents to the second term from the left in the expression we got in the last step, then we'll simplify the resulting expression,

Let's summarize the solution steps:

$4^5-4^6\cdot\frac{1}{4}= 4^5-4^6\cdot4^{-1} =4^5-4^{5}=0$

We got that the answer is 0,

Therefore the correct answer is answer A.

We'll use the power rule for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$and we'll simplify the second term on the left in the equation using it:
$5^3+5^{-3}\cdot5^3=5^3+5^{-3+3}=5^3+5^0=5^3+1$ where in the first stage we applied the mentioned rule to the second term on the left, then we simplified the expression with the exponent, and in the final stage we used the fact that any number raised to the power of 0 equals 1,

We didn't touch the first term of course since it was already simplified,

Therefore the correct answer is answer C.

Recall the law of exponents for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

We'll use this to deal with the fraction's denominator in the problem:

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=\frac{(-3)^5\cdot8^4}{(-3)^{3+2+(-5)}}=\frac{(-3)^5\cdot8^4}{(-3)^0}$

In the first stage, we'll apply the above law to the denominator and then proceed to simplify the expression with the exponent in the denominator.

Remember that raising any number to the power of 0 gives the result 1, or mathematically:

$X^0=1$

Therefore the denominator that we obtain in the last stage is 1.

This means that:

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=\frac{(-3)^5\cdot8^4}{(-3)^0}=\frac{(-3)^5\cdot8^4}{1}=(-3)^5\cdot8^4$

Recall the law of exponents for an exponent of a product inside of parentheses is as follows:

$(x\cdot y)^n=x^n\cdot y^n$

Apply this law to the first term in the product:

$(-3)^5\cdot8^4=(-1\cdot3)^5\cdot8^4 =(-1)^5\cdot3^5\cdot8^4=-1\cdot 3^5\cdot 8^4=-3^5\cdot8^4$

Note that the exponent applies separately to both the number 3 and its sign, which is the minus sign that is in fact multiplication by $-1$.

Let's summarize everything we did:

$\frac{(-3)^5\cdot8^4}{(-3)^3(-3)^2(-3)^{-5}}=(-3)^5\cdot8^4 = -3^5\cdot8^4$

Therefore the correct answer is answer C.

In order to solve the given problem, we will follow these steps:

• Step 1: Simplify $2^0$. According to the zero exponent rule, $2^0 = 1$.

• Step 2: Simplify $3^{-4}$. Using the negative exponent rule, $3^{-4} = \frac{1}{3^4}$.

• Step 3: Simplify $9^2$. Recognize that $9 = 3^2$, thus $9^2 = (3^2)^2 = 3^{4}$.

• Step 4: Substitute the simplified terms back into the expression:

$\frac{2^0 \cdot 3^{-4}}{5^4 \cdot 9^2} = \frac{1 \cdot \frac{1}{3^4}}{5^4 \cdot 3^{4}}$

• Step 5: Simplify by combining like bases: since $3^{-4}$ in the numerator can be combined with $3^4$ in the denominator, you have:

$= \frac{1}{5^4 \cdot 3^{4+4}} = \frac{1}{5^4 \cdot 3^8}$

Therefore, the simplified expression is $\frac{1}{5^4 \cdot 3^8}$.

The problem requires simplification $\frac{9^2 \cdot 3^{-4}}{6^3}$ using exponent rules. Here’s a step-by-step guide to solving it:

• Step 1: Convert each term to powers of a common base.

  Notice that $9$ is $3^2$ and $6$ is $2 \times 3$. Hence:

  $9^2 = (3^2)^2 = 3^{4}$

  Therefore, the expression becomes $\frac{3^4 \cdot 3^{-4}}{(2 \cdot 3)^3}$.

• Step 2: Simplify the numerator.

  Using the exponent multiplication rule: $3^4 \cdot 3^{-4} = 3^{4 + (-4)} = 3^0 = 1$.

• Step 3: Expand the denominator.

  Calculate $(2 \cdot 3)^3$ by applying the distributive property: $(2 \cdot 3)^3 = 2^3 \cdot 3^3$.

• Step 4: Simplify the expression.

  After simplifying, the entire expression is $\frac{1}{2^3 \cdot 3^3}$.

  This simplifies further to $\frac{1}{6^3} = 6^{-3}$, because $2^3 \cdot 3^3 = (2 \cdot 3)^3 = 6^3$.

Therefore, the solution to the problem is $6^{-3}$.

Let's start by simplifying the second term in the complete multiplication, meaning - the fraction. We'll simplify it in two stages:

In the first stage we'll use the power law for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

and simplify the fraction's numerator:

$\frac{19^{35}\cdot19^{-32}}{19^3}=\frac{19^{35+(-32)}}{19^3}=\frac{19^{35-32}}{19^3}=\frac{19^3}{19^3}$

Next, we can either remember that dividing any number by itself gives 1, or use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$ to get that:$\frac{19^3}{19^3}=19^{3-3}=19^0=1$

where in the last step we used the fact that raising any number to the power of 0 gives 1, meaning mathematically that:

$X^0=1$

Let's summarize this part, we got that:

$\frac{19^{35}\cdot19^{-32}}{19^3}=1$

Let's now return to the complete expression in the problem and substitute this result in place of the fraction:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}\cdot1=3^{-3}$

In the next stage we'll recall the power law for negative exponents:

$a^{-n}=\frac{1}{a^n}$

and apply this law to the result we got:

$3^{-3}=\frac{1}{3^3}=\frac{1}{27}$

Summarizing all the steps above, we got that:

$3^{-3}\cdot\frac{19^{35}\cdot19^{-32}}{19^3}=3^{-3}=\frac{1}{27}$

Therefore the correct answer is answer A.

To solve this problem, we'll follow these steps:

• Step 1: Simplify $\sqrt[6]{b^{12}}$ using fractional exponent form.
• Step 2: Simplify the expression $\left(\frac{1}{b}\right)^2$ using negative exponents.
• Step 3: Combine and simplify the entire expression.

Now, let's work through each step:

Step 1: Simplify $\sqrt[6]{b^{12}}$.
The expression $\sqrt[6]{b^{12}}$ can be rewritten using fractional exponents as $b^{12/6} = b^2$.

Step 2: Simplify $\left(\frac{1}{b}\right)^2$.
The expression $\left(\frac{1}{b}\right)^2$ simplifies using negative exponents: $b^{-2}$.

Step 3: Combine the simplified expressions and the original variable $a$.
Combine all components as follows:
$b^2 \cdot b^{-2} \cdot a$.
Using the property $x^m \cdot x^n = x^{m+n}$, we have:
$b^{2 + (-2)} \cdot a = b^0 \cdot a$.
Since $b^0 = 1$ (by the zero exponent rule), the expression simplifies to:
$a$.

Therefore, the solution to the problem is $a$.

We use the power property to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Keep in mind that this property is also valid for several terms in the multiplication and not just for two, for example for the multiplication of three terms with the same base we obtain:

$a^m\cdot a^n\cdot a^k=a^{m+n}\cdot a^k=a^{m+n+k}$When we use the mentioned power property twice, we could also perform the same calculation for four terms of the multiplication of five, etc.,

Let's return to the problem:

$$$$Keep in mind that all the terms of the multiplication have the same base, so we will use the previous property:

$5^{-3}\cdot5^0\cdot5^2\cdot5^5=5^{-3+0+2+5}=5^4$Therefore, the correct answer is option c.

Note:

Keep in mind that $5^0=1$